1 / 26

 Water is polar and forms H-bonds with itself responsible for high boiling point of water

d -. d +. d +. O. H. H. Na +. H. Cl -. H. O. O. H. H. Water: The Medium of Life.  Water is polar and forms H-bonds with itself responsible for high boiling point of water  Water forms H-bonds with polar and ionic substances

thaddeus-oneill
Download Presentation

 Water is polar and forms H-bonds with itself responsible for high boiling point of water

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. d- d+ d+ O H H Na+ H Cl- H O O H H Water: The Medium of Life Water is polar and forms H-bonds with itself responsible for high boiling point of water Water forms H-bonds with polar and ionic substances responsible for solubility of ions and molecules Substance MW BP (°C) Water (H2O) 18.02 100.0 Ammonia (NH3) 17.03 -33.4 Methane (CH4) 16.04 -161.5 H H O H H O O H H Ion-dipole interactions dipole-dipole interactions

  2. Equilibrium Reactions Reactions do not go to completion Rate of forward reaction equals rate of reverse reaction Equilibrium reactions can be described by a mathematical equation A + B C + D products The greater degree of the forward reaction the bigger Keq reactants At equilibrium, the rate of forward = rate of reverse so ratio of products:reactants is constant – therefore, Keq is constant.

  3. H2O H+ + OH- H+ + H2O H3O+ For pure water at 25 °C, 1 atm: Water self-dissociates equilibrium constant Hydrogen ion often called a proton Hydroxide ion (base) @25 C, pure H2O is 55.5 M - constant Hydronium ion (acid) x(x)=1x10-14 x2=1x10-14 x=1x10-7 Ion-product constant of water

  4. At 25 C, 1atm, Kw always = 10-14 so you can always calculate H+ or OH- concentration if given the other Example: Aqueous solution has H+ concentration of 10-2 M, what is OH- concentration? Kw=1x10-14=[H+][OH-]=1x10-2 [OH-] therefore, [OH-] = 1x10-12

  5. Can act as an acid or base Water is amphoteric (10.4) Brønstead-Lowry definitions Acid - substance that gives H+ to another molecule or ion Base - substance that accepts H+ from an acid (uses lone pair electrons) Is water an acid or base in the following reactions? H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+ (aq) F-(aq) + H2O(l) HF (aq) + OH-(aq) NH4 +(aq) + H2O(l) NH3 (aq) + H3O+ (aq) Base Acid Base

  6. H-A + H2O H3O+ + A- 1 10-1 10-2 10-3 10-4 10-5 10-6 10-8 10-9 10-10 10-11 10-12 10-13 10-14 10-7 8 0 1 9 10 2 11 3 12 4 13 5 14 6 figure 10.2 on page 276 Measuring Acidity  A solution’s acidity depends on the concentration of H3O+ The higher the conc of H+, the more acidic the solution  pH scale gives a convenient comparison of acidity H+ concentration ranges over 14 fold magnitude which solution is more acidic? [H3O+] = 9.0 x 10-8 M or [H3O+] = 3.5 x 10-7 M which solution is more acidic? pH = 7.05 or pH = 6.46 [H3O+] pH 7 neutral acidity basicity

  7. assign application on page 280 of text for reading pH example problems (use equation): 1. The H+ concentration in coffee is about 1 x 10-5 M. What pH is this? 2. Soft drinks usually have a pH of approximately 3.1. What is the [H3O+] in a soft drink? 3. A cleaning solution was found to have [OH-]=1 x 10-3 M. What is the pH? A logarithim contains the same number of digits to the right of the decimal point that the original number has. An antilogarithm contains the same number of digits that the original number has to the right of the decimal point.

  8. Acids Produce Conjugate Bases (4.12, 6.10, Chap.10) H-A + H2O H3O+ + A- …then this is a weak base because it has little affinity for a proton. If this is a strong acid because it gives up a proton readily... conjugate acid conjugate base acid base H-A + H2O H3O+ + A- …then this is a strong base because it has a high affinity for a proton. If this is a weak acid because it gives up a proton with difficulty...  Strong acids give up hydrogens easily and produce weak conjugate bases Dissociation - splitting of an acid into hydrogen ion and an anion Strong acids dissociate “100%”  Weak acids don’t give up hydrogens easily and produce strong conjugate bases Weak acids dissociate much less than “100%” conjugate acid conjugate base acid base The stronger the acid, the weaker its conjugate base. The weaker the acid, the stronger its conjugate base.

  9. H-A + H2O H3O+ + A- - + [ A ][ H O ] = 3 K eq [ HA ] [ H O ] 2 - + [ A ][ H O ] = = 3 K [ H O ] K eq 2 a [ HA ] Acid Dissociation Constant, Ka  Dissociation of a weak acid is an equilibrium rxn General rxn for dissociation of a weak acid:  Equilibrium constant reveals extent of rxn (7.7, 10.7-10.8) Write equilibrium equation for the dissociation of a weak acid: Concentration of water is essentially constant so a new constant is defined - the acid dissociation constant, Ka

  10. H-A + H2O H3O+ + A-  Weak acids have Ka << 1  pKa scale gives a convenient comparison of acid strength - + If the acid doesn’t dissociate very much, HA is large. The larger the denominator, the smaller Ka. [ A ][ H O ] = 3 K a [ HA ] As [H+] , pH  As [OH-] , pOH  As Ka , pKa  - This means that strong weak acids (those with large Ka because HA is small) have low pKa and weak weak acids (those with small Ka because HA is large have high pKa

  11. Acid Dissociation Constants of Common Weak Acids

  12. Ka and pKa example problems (use equation): 1. Place the following acids in order of increasing strength based on their pKa values: Acid pKa H2CO3 6.37 NH4+ 9.25 HF 3.45 CH3COOH 4.75 H3PO4 2.12 2. At equilibrium, the conjugate base of formic acid has a concentration of .0029 M. What is the concentration of formic acid present at equilibrium? 3. At equilibrium, an unknown weak monoprotic acid has a concentration of 0.1 M and the concentration of conjugate base is .0037 M. What is its pKa and what is the acid?

  13. Acid-Base Titration (10.15)  Used to determine concentration of an acid by addition of known amounts of base (usually NaOH) using a color indicator, base is added until neutralization is complete. Volume and molarity of base is used to determine original concentration of acid Example: To determine the concentration of the acid in an old bottle of HCL whose label had become unreadable, a titration was carried out. What is the HCL concentration if 58.4 ml of 0.250 M NaOH was required to titrate a 20.0 ml sample of the acid?

  14. Titration Curve  Used to identify an unknown weak acid through its pKa pH is monitored as known volumes of base are added to the acid solution. Discuss starting pH, equivalence point, and buffer region Equivalence point Half-equivalence point (50.00/2) ml = 25.0 ml 50.00 ml

  15. H-A H3O+ + A- [ HA ] [ HA ] + + - = = - log[ [ H H O O ] ] K log( K ) 3 a 3 a - - [ A ] [ A ] [ HA ] = - - pH log K log( ) a - [ A ] - [ A ] [ HA ] = + = - pH pK log( ) pH pK log( ) a a - [ HA ] [ A ]  Mathematical relationship of pH and pKa Henderson-Hasselbalch equation: equation relates pH and Ka Rearrange for [H3O+]: Take the negative log of both sides: Recall: log(xy) = log(x) + log(y) and log(x/y)=log(x) - log(y) Split up right side of equation (be sure to distribute the negative sign) and rearrange: When A-=HA, pH=pKa - very important for buffering OR When A-=HA, pH=pKa

  16.  Experimentally determining identity of acid through pKa Titration of ________________ with NaOH Step 3 Step 1 Step 2 Step 1: find the equivalence point Step 2: Take 1/2 volume needed to reach equivalence point - here, [HA]=[A-] Step 3: determine pH at where [HA]=[A -] pH = pKa at 1/2 equivalence point

  17. 2nd half-equivalence 2nd Equivalence point 1st half-equivalence 1st Equivalence point  diprotic acids have biphasic curves

  18. Draw a model titration curve for a triprotic acid

  19. Buffers and pKa (10.12)  Buffers maintain pH Constant pH crucial for proper organ function enzymes protein structure respiration

  20.  A buffer is a solution of a weak acid and its conjugate base weak acid can neutralize moderate amounts of added base conjugate base can neutralize moderate amounts of added acid Example above: Mixture of CH3COOH and CH3COO- CH3COOH + OH- CH3COO- + H2O pH rises slightly CH3COO- + H3O+ CH3COOH + H2O pH lowers slightly Buffer capacity is maximal when conc. of weak acid = conc of conjugate base Buffer capacity is maximal when pH=pKa

  21. H2CO3 + H2O HCO3-2 + H3O+ Major Physiological Buffers (10.13)  dihydrogen phosphate-hydrogen phosphate buffer system  carbonic acid – bicarbonate buffer system pKa = 7.2 H2PO4- + H2O HPO4-2 + H3O+ Most important intracellular buffer. Phosphate concentration too low extra- cellular to be a sufficient buffer pKa = 6.37 Most important buffer in blood. But blood pH is 7.4 so this buffer is near the end of the buffer range of this system. Allows more buffering effect for added acid than for added base – excessive acid production is more common in the body than excessive loss of acid or addition of base. Carbonic acid is unstable and breaks down into CO2 and H2O H2CO3 CO2 + H2O makes blood pH intimately related to breathing rates

  22. CO2 + H2O H2CO3 HCO3-2 + H3O+ [ CO ] + = 2 [ H O ] K 3 a - [ HCO ] 3 - + [ HCO ][ H O ] = 3 3 K a [ CO ] 2 Increase here causes lower pH decrease here causes higher pH

  23. Acidosis Questions. Explain in detail with structures and equations where applicable. 1. Stroke and heart attack patients often experience acidosis. Explain in detail. 2. Although not recommended, some sprinters hyperventilate for 30 seconds just prior to a race. Why is this supposed to work? 3. Asthma and emphysema patients are at risk for developing acidosis. Explain in detail. 4. Why does exposure to high altitudes temporarily induces mild alkalosis? Lack of blood flow causes lack of oxygen and increased production of lactic acid which accumulates in blood causes drop in pH. Sprinting produces excess lactic acid due to insufficient oxygen. To counteract the lactic acid, sprinters may hyperventilate to induce a temporary state of alkalosis. Difficulty breathing causes lack of oxygen and increased production of lactic acid which accumulates in blood causes drop in pH. Similar to hyperventilation. The atmosphere causes respiration to increase which causes more carbon dioxide to be expired. This causes a shift towards H2CO3 lowering the conc. of H+ and increasing pH.

  24. ammonia Primary amine (1° amine) Secondary amine (2° amine) Tertiary amine (3° amine) + H3O+ + H2O acid ammonium ion 1° amine RNH3+ + OH- RNH2 + H2O Amines Are Organic Bases  Amines are ammonia derivatives  Amines are weak bases Ammonia and amines are weak bases because its lone pairs can accept a proton from an acid or water General Structure: Ammonium ions have pKa ~10 (They are 50% unprotonated around pH 10) pH = 4 - protonated pH = 7 - protonated pH = 11 - unprotonated

  25. General Structure + H2O + H3O+ Pentanoate ion Pentanoic acid Carboxylic Acids Are Organic Acids Examples: Common NameFormal Name pKa Formic acid (ants) Methanoic acid 3.75 Acetic acid (vinegar) Ethanoic acid 4.76 Butyric acid (rancid butter) Butanoic acid 4.81 Caproic acid (goat smell) Hexanoic acid 4.84 Carboxylic acids have pKa ~4-5 (They are 50% unprotonated around pH 4-5) pKa of pentanoic acid is 4.82 pH = 4 - mostly protonated pH = 7 - mostly unprotonated

  26. Using pKa charts, draw the predominant structure for the following parent compounds at the pH indicated Parent CompoundpKaStructure at pH: pka=4.87 pH = 7 Propanoic acid pka1=2.9 pka2=5.4 pH = 4 Phthalic acid pH = 7 pka=10.72 Dimethylammonium ion pH = 11 pka=9.3 Cyclohexylammonium ion pH = 6 pka=9.7 Trimethylammonium ion pH = 7 Alanine

More Related